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I am curious about the submodules of a module with a given property. Let $M$ be an $R$-module.

Any advice is helpful. Thank you.

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For any non-noetherian ring, there is a finitely generated $R$-module, which has a non-finitely generated submodule, namely $R$ itself. So there is no generalization.

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For #1: no. For example, every ring $R$ is a finitely generated module over itself (namely, it is generated by $1$). However, the submodules of $R$ correspond with ideals of $R$, so if $R$ is not noetherian, then there is some ideal which is not finitely generated. In the case $R = \mathbb{Z}$, this is true; I refer you to this post.

For #2: once again, no. Consider $R$ as a free $R$-module. Then again, the submodules of $R$ are ideals. If $I$ is a free $R$-module, then it has a $R$-basis; but you can see that any two elements of an ideal $I$ are linearly dependent, so any ideal which is a free $R$-module has to be principal. Now just take $R$ to be a ring which is not a PID, e.g. $R = \mathbb{Z}[x]$. However, if $R$ is a PID, then this is true (in particular, this resolves the case $R = \mathbb{Z}$). This is a good exercise, so you should try it yourself. If you can't figure it out, feel free to comment, and I will post a proof or refer you to one on the web.

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It depends on the ring you're working on. If $R$ is Noetherian, then a submodule of a finitely generated module is always finitely generated. So this works for $\Bbb Z$ in particular. For a counter-example, take $R=k[x_1,x_2,...]$ (which isn't Noetherian). Then $(x_1,x_2,...)\subset R$ is not finitely generated, but $R$ is (as a $R$-module).

For your second question, again it depends on the ring. This is true for PIDs, in particular for $\Bbb Z$.

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